• Feb 6th 2009, 10:44 AM
Moo
Hi

I'd like to know something... Let's assume I'm given a function ($\displaystyle f(x_1,x_2)=\dots$). I'm then asked to prove that this function is a probability density function.

So I have to check that $\displaystyle \iint_{A \times B} f(x_1,x_2) ~dx_1dx_2=1$
Is it enough ?

And do you guys have to prove first that the function is integrable ? Or do you even prove that it's integrable ?

These stats courses are driving me nuts (Rofl)
• Feb 6th 2009, 11:29 AM
Laurent
Quote:

Originally Posted by Moo
Hi

I'd like to know something... Let's assume I'm given a function ($\displaystyle f(x_1,x_2)=\dots$). I'm then asked to prove that this function is a probability density function.

So I have to check that $\displaystyle \iint_{A \times B} f(x_1,x_2) ~dx_1dx_2=1$
Is it enough ?

And do you guys have to prove first that the function is integrable ? Or do you even prove that it's integrable ?

These stats courses are driving me nuts (Rofl)

Hi,

first of all $\displaystyle f$ has to be non-negative.

Then, integrating $\displaystyle f$ must make sense: i.e., depending on the level of study, $\displaystyle f$ has to be, say, piecewise continuous (for simple Riemann integration) or Lebesgue measurable. This part is usually skipped because it is readily checked.

And finally, the integral of $\displaystyle f$ on its domain must equal 1. Note that it is not mandatory to first prove that the integral is finite. Indeed $\displaystyle f$ is non-negative, so that the integral always has a value, either finite or infinite, and the computation will show that it is 1.

Good luck with your probability/stats course, it's worth it!
• Feb 6th 2009, 11:53 AM
Moo
Quote:

Originally Posted by Laurent
Hi,

first of all $\displaystyle f$ has to be non-negative.

Then, integrating $\displaystyle f$ must make sense: i.e., depending on the level of study, $\displaystyle f$ has to be, say, piecewise continuous (for simple Riemann integration) or Lebesgue measurable. This part is usually skipped because it is readily checked.

And finally, the integral of $\displaystyle f$ on its domain must equal 1. Note that it is not mandatory to first prove that the integral is finite. Indeed $\displaystyle f$ is non-negative, so that the integral always has a value, either finite or infinite, and the computation will show that it is 1.

Arf this is indeed simple.
Hmmm how would you translate in French "integration makes sense" ? Is it just proving that it's Lebesgue measurable ?

erm... I'll continue on measures... How can one prove that a function is measurable (with respect to any measure - Lebesgue, or counting, mostly) ?
And is there an equivalence between a function being continuous and the function being measurable ?

Quote:

Good luck with your probability/stats course, it's worth it!
Haha I must trust you on it :D
I've got 3 courses of probability & stats (Rofl)

Thanks =)
• Feb 6th 2009, 01:10 PM
Laurent
Quote:

Originally Posted by Moo
Arf this is indeed simple.
Hmmm how would you translate in French "integration makes sense" ? Is it just proving that it's Lebesgue measurable ?

I wasn't sure whether you knew measure theory; then I indeed meant "Lebesgue measurable".
Otherwise, Riemann integration suffices for piecewise continuous density functions (and a few others), like the usual ones.

Quote:

erm... I'll continue on measures... How can one prove that a function is measurable (with respect to any measure - Lebesgue, or counting, mostly) ?
And is there an equivalence between a function being continuous and the function being measurable ?
Measurability is actually not defined with respect to a measure but to a $\displaystyle \sigma$-field ("une tribu"). I shouldn't have said "Lebesgue measurable", it was a slight mistake... (Or I could pretend I was talking about the Lebesgue $\displaystyle \sigma$-field)

The Lebesgue measure is defined on the Borel $\displaystyle \sigma$-field ("la tribu des boréliens"). Continuous functions are measurable with respect to this $\displaystyle \sigma$-field, but there are plenty of other measurable functions. So many that it is somewhat difficult to find a non-measurable one! In fact it is not possible to build non-measurable functions (with respect to the borelian $\displaystyle \sigma$-field) without using the axiom of choice. In practice, any function is measurable, except if it was meant to be a counter-example to this property (Wink)

The counting measure is defined on the discrete $\displaystyle \sigma$-field, for which any function is measurable.
• Feb 7th 2009, 05:30 AM
Moo
Quote:

Originally Posted by Laurent
I wasn't sure whether you knew measure theory; then I indeed meant "Lebesgue measurable".
Otherwise, Riemann integration suffices for piecewise continuous density functions (and a few others), like the usual ones.

Measurability is actually not defined with respect to a measure but to a $\displaystyle \sigma$-field ("une tribu"). I shouldn't have said "Lebesgue measurable", it was a slight mistake... (Or I could pretend I was talking about the Lebesgue $\displaystyle \sigma$-field)

The Lebesgue measure is defined on the Borel $\displaystyle \sigma$-field ("la tribu des boréliens"). Continuous functions are measurable with respect to this $\displaystyle \sigma$-field, but there are plenty of other measurable functions. So many that it is somewhat difficult to find a non-measurable one! In fact it is not possible to build non-measurable functions (with respect to the borelian $\displaystyle \sigma$-field) without using the axiom of choice. In practice, any function is measurable, except if it was meant to be a counter-example to this property (Wink)

The counting measure is defined on the discrete $\displaystyle \sigma$-field, for which any function is measurable.

Okay, it all looks clear (Nod)

So basically, we just have to check that :
- the function is measurable
- the unfction is positive
- the integral is 1, there is no need to prove that the integral is finite.

One last question : in order to prove that a function is the pdf, is it necessary to check the other axioms of a pdf (continuity from the right, limits at + and - infinity, etc...) ?
Maybe I'm just getting confused lol... the lessons are not acquired yet ><
• Feb 7th 2009, 07:01 AM
Laurent
Quote:

Originally Posted by Moo
One last question : in order to prove that a function is the pdf, is it necessary to check the other axioms of a pdf (continuity from the right, limits at + and - infinity, etc...) ?
Maybe I'm just getting confused lol... the lessons are not acquired yet ><

To prove that a function $\displaystyle f$ is a p.d.f., depending on how it is defined, either you can simply write it as the pdf of some r.v. $\displaystyle X$: $\displaystyle f(x)=P(X\leq x)$ for every $\displaystyle x\in\mathbb{R}$, otherwise you have to check that $\displaystyle f$ is increasing, right-continuous, and with respective limits 0 and 1 at $\displaystyle -\infty$ and $\displaystyle +\infty$, that's all.